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avatar+322 

\(\lim_{n\rightarrow ∞} \sum_{i=1}^{n} \frac{2}{n}(\frac{i}{n})^2\)

 

Just to make things easier↓↓

\(i^2\) = \(\frac{n(n+1)(2n+1)}{6}\)

 Apr 18, 2019
 #1
avatar+7685 
+2

It is easy enough, even with that incorrect formula :)

The correct formula should be \(\displaystyle \sum^{n}_{i=1} i^2 = \dfrac{n(n+1)(2n+1)}{6}\).

Obviously, this limit is a Riemann integral.

\( S = \displaystyle{\lim_{n\rightarrow \infty} \sum_{i=1}^{n}\left(\left(\dfrac{2}{n}\right)\left(\dfrac{i}{n}\right)^2\right)\\ \;\;\;\!= 2\lim_{n\rightarrow \infty} \sum_{k=1}^{n}\left(\left(\dfrac{1}{n}\right)\left(\dfrac{k}{n}\right)^2\right)\\ \boxed{\text{Riemann integral: For any function f continuous on [0,1], } \lim_{n\rightarrow \infty}\sum^{n}_{k=1}\dfrac{1}{n}f\left(\dfrac{k}{n}\right) = \int^1_0f(x)dx }\\ S=2\int^1_0x^ 2dx\\\;\;\;\!=2\left(\dfrac{1}{3}\right)\\\;\;\;\!=\dfrac{2}{3} }\)

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 Apr 18, 2019

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